3.34 \(\int \frac {(A+B x) (b x+c x^2)^3}{x^3} \, dx\)

Optimal. Leaf size=38 \[ \frac {B (b+c x)^5}{5 c^2}-\frac {(b+c x)^4 (b B-A c)}{4 c^2} \]

[Out]

-1/4*(-A*c+B*b)*(c*x+b)^4/c^2+1/5*B*(c*x+b)^5/c^2

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Rubi [A]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {765} \[ \frac {B (b+c x)^5}{5 c^2}-\frac {(b+c x)^4 (b B-A c)}{4 c^2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^3)/x^3,x]

[Out]

-((b*B - A*c)*(b + c*x)^4)/(4*c^2) + (B*(b + c*x)^5)/(5*c^2)

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x^3} \, dx &=\int \left (\frac {(-b B+A c) (b+c x)^3}{c}+\frac {B (b+c x)^4}{c}\right ) \, dx\\ &=-\frac {(b B-A c) (b+c x)^4}{4 c^2}+\frac {B (b+c x)^5}{5 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 67, normalized size = 1.76 \[ A b^3 x+\frac {1}{2} b^2 x^2 (3 A c+b B)+\frac {1}{4} c^2 x^4 (A c+3 b B)+b c x^3 (A c+b B)+\frac {1}{5} B c^3 x^5 \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^3)/x^3,x]

[Out]

A*b^3*x + (b^2*(b*B + 3*A*c)*x^2)/2 + b*c*(b*B + A*c)*x^3 + (c^2*(3*b*B + A*c)*x^4)/4 + (B*c^3*x^5)/5

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fricas [B]  time = 0.81, size = 69, normalized size = 1.82 \[ \frac {1}{5} \, B c^{3} x^{5} + A b^{3} x + \frac {1}{4} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{4} + {\left (B b^{2} c + A b c^{2}\right )} x^{3} + \frac {1}{2} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^3,x, algorithm="fricas")

[Out]

1/5*B*c^3*x^5 + A*b^3*x + 1/4*(3*B*b*c^2 + A*c^3)*x^4 + (B*b^2*c + A*b*c^2)*x^3 + 1/2*(B*b^3 + 3*A*b^2*c)*x^2

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giac [B]  time = 0.16, size = 72, normalized size = 1.89 \[ \frac {1}{5} \, B c^{3} x^{5} + \frac {3}{4} \, B b c^{2} x^{4} + \frac {1}{4} \, A c^{3} x^{4} + B b^{2} c x^{3} + A b c^{2} x^{3} + \frac {1}{2} \, B b^{3} x^{2} + \frac {3}{2} \, A b^{2} c x^{2} + A b^{3} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^3,x, algorithm="giac")

[Out]

1/5*B*c^3*x^5 + 3/4*B*b*c^2*x^4 + 1/4*A*c^3*x^4 + B*b^2*c*x^3 + A*b*c^2*x^3 + 1/2*B*b^3*x^2 + 3/2*A*b^2*c*x^2
+ A*b^3*x

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maple [B]  time = 0.05, size = 73, normalized size = 1.92 \[ \frac {B \,c^{3} x^{5}}{5}+A \,b^{3} x +\frac {\left (A \,c^{3}+3 B b \,c^{2}\right ) x^{4}}{4}+\frac {\left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{3}}{3}+\frac {\left (3 A \,b^{2} c +b^{3} B \right ) x^{2}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^3/x^3,x)

[Out]

1/5*B*c^3*x^5+1/4*(A*c^3+3*B*b*c^2)*x^4+1/3*(3*A*b*c^2+3*B*b^2*c)*x^3+1/2*(3*A*b^2*c+B*b^3)*x^2+A*b^3*x

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maxima [B]  time = 0.90, size = 69, normalized size = 1.82 \[ \frac {1}{5} \, B c^{3} x^{5} + A b^{3} x + \frac {1}{4} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{4} + {\left (B b^{2} c + A b c^{2}\right )} x^{3} + \frac {1}{2} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^3,x, algorithm="maxima")

[Out]

1/5*B*c^3*x^5 + A*b^3*x + 1/4*(3*B*b*c^2 + A*c^3)*x^4 + (B*b^2*c + A*b*c^2)*x^3 + 1/2*(B*b^3 + 3*A*b^2*c)*x^2

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mupad [B]  time = 0.03, size = 65, normalized size = 1.71 \[ x^2\,\left (\frac {B\,b^3}{2}+\frac {3\,A\,c\,b^2}{2}\right )+x^4\,\left (\frac {A\,c^3}{4}+\frac {3\,B\,b\,c^2}{4}\right )+\frac {B\,c^3\,x^5}{5}+A\,b^3\,x+b\,c\,x^3\,\left (A\,c+B\,b\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^3*(A + B*x))/x^3,x)

[Out]

x^2*((B*b^3)/2 + (3*A*b^2*c)/2) + x^4*((A*c^3)/4 + (3*B*b*c^2)/4) + (B*c^3*x^5)/5 + A*b^3*x + b*c*x^3*(A*c + B
*b)

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sympy [B]  time = 0.08, size = 73, normalized size = 1.92 \[ A b^{3} x + \frac {B c^{3} x^{5}}{5} + x^{4} \left (\frac {A c^{3}}{4} + \frac {3 B b c^{2}}{4}\right ) + x^{3} \left (A b c^{2} + B b^{2} c\right ) + x^{2} \left (\frac {3 A b^{2} c}{2} + \frac {B b^{3}}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**3/x**3,x)

[Out]

A*b**3*x + B*c**3*x**5/5 + x**4*(A*c**3/4 + 3*B*b*c**2/4) + x**3*(A*b*c**2 + B*b**2*c) + x**2*(3*A*b**2*c/2 +
B*b**3/2)

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